Forum:Are there ordinal hierarchies without fundamental sequences?
Define an ordinal hierarchy as \(F: \omega_1 \mapsto (\mathbb{N} \mapsto \mathbb{N})\) where for all \(\alpha > \beta\), \(F(\alpha)\) eventually outgrows \(F(\beta)\). Define a fundamental sequence system as \(S: \omega_1 \mapsto (\omega_1)^\omega\) (where \((\omega_1)^\omega\) is the set of all increasing \(\omega\)-sequences in \(\omega_1\)) such that for all limit ordinals \(\alpha\) we have \(\sup\{\beta: \beta \in S(\alpha)\} = \alpha\). If we take the axiom of choice, ordinal hierarchies and fundamental sequence systems must exist, so let's scale it back to ZF. Is there a model of ZF where ordinal hierarchies exist but fundamental sequence systems don't? it's vel 07:21, October 4, 2014 (UTC) I've shown that it's impossible. I'll give a proof when I'm back home. LittlePeng9 (talk) 07:27, October 4, 2014 (UTC) I take that claim back for now. LittlePeng9 (talk) 07:36, October 4, 2014 (UTC) Just a remark, we actually need some of axiom of choice for any of these to exist. This is because we cannot prove in ZF alone that there is a subset of \(\Bbb R\) which has cardinality \(\omega_1\) (same applies to \(\Bbb N^\Bbb N\)). Ordinal hierarchy necessarilly gives us an injection from \(\omega_1\) to \(\Bbb N^\Bbb N\). For existence of fundamental sequence this is less straightforward, but we can for every \(\alpha<\omega_1\) construct a real number \(r_\alpha\) which codes \(\alpha\), by using \(S\) function. LittlePeng9 (talk) 12:22, October 4, 2014 (UTC) Here's a partial answer. The SGH, HH, and FGH appear to have the following property: for every limit ordinal \(\alpha\), there does not exist a function f such that \(f(n) > F_{\alpham}(n)\) for all \(m,n \in \Bbb N\) (here F is the ordinal hierarchy in question). We can turn this into a version without fundamental sequences: for every limit ordinal \(\alpha\) and every function f on \(\Bbb N\), there exists a \(\beta < \alpha\) such that, if \(f(n) > F_{\gamma} (n)\) for all \(n \in \Bbb N\), then \(\gamma < \beta\). We will also assume that each \(F_\alpha\) is weakly increasing. If an ordinal hierarchy F on \(\omega_1\) has these properties, then we can define fundamental sequences for all countable ordinals by letting \(\alphan\) be the supremum of all ordinals \(\beta\) such that \(f(m) > F_{\beta}(m)\) for all \(m \ge n\). By the previous property, \(\alphan\) will be less than \(\alpha\) (if \(\alphan = \alpha\) then by setting \(f(m) = f(n)\) for \(m < n\) we can insure that f dominates \(F_\alpha\) for a cofinal subset of \(\alpha\)), and the fact that \(F_\alpha\) eventually dominates \(F_\beta\) for all \(\beta < \alpha\) insures that \(\sup \alphan = \alpha\). So we get fundamental sequences if we have an F with these properties. Deedlit11 (talk) 14:16, October 4, 2014 (UTC) : Your first claim seems to be correct for standard hierarchies, because almost always we have that \(F_{\alphan}(2)\), as a function of \(n\), is unbounded. These functions however require the fundamental sequences to be specified, so it doesn't always have to be the case. There are some trivial examples for which it doesn't hold, however: define \(F_i(n)\) to be \(0\) if \(n